Question

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Answer 1

$\frac{ { \sec }^{2}(90 - \alpha) - { \cot}^{2} \alpha }{2( { \sin }^{2}25 + { \sin }^{2}65) } + \frac{2 { \cos}^{2}60. { \tan}^{2}28 .{ \tan}^{2} 62 }{ \sin30 . \cos60 } \\ \\ = > \frac{ { \csc}^{2} \alpha - { \cot }^{2} \alpha }{2( { \sin(90 - 65) }^{2} + { \sin }^{2}65) } + \frac{2 { \cos }^{2}60. { \tan(90 - 62) }^{2} . { \tan}^{2}62 }{ \sin30. \cos60 } \\ \\ = > \frac{1}{2( { \cos}^{2}65 + { \sin }^{2} 65)} + \frac{2 { \cos }^{2}60. { \cot}^{2} 62. { \tan }^{2} 62}{ \sin30. \cos60 } \\ \\ = > \frac{1}{2 \times 1} + \frac{2 \times { (\frac{1}{2}) }^{2} \times 1}{ \frac{1}{2} \times \frac{1}{2} } \\ \\ = > \frac{1}{2} + 2 \\ \\ = > \frac{5}{2}$

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Answer 2

Step-by-step explanation:

answer : 1.6 × 10^25 MeV

Let's first find binding energy of each copper nucleus and then we can easily find binding energy of 3.0 g of ^{63}_{29}Cu

29

63

Cu .

we know,

mass of each proton = 1.00783u

mass of each neutron = 1.00867u

in ^{63}_{29}Cu

29

63

Cu presents 29 protons and (63 - 29) = 34 neutrons.

now mass of 29 protons = 29 × 1.00783 = 29.22707u

mass of 34 neutrons = 34 × 1.00867u

= 34.29478u

Total theoretical mass = 29.22707u + 34.29478u = 63.52185u

mass defect , ∆m = 63.52185u - 62.92960u = 0.59225u

so, binding energy of each Cu nucleus = ∆m × 931MeV

= 0.59225 × 931 = 551.385MeV

let's find number of cu atoms in 3g .

number of cu atoms = 3g/63g × 6.023 × 10²³ = 2.86 × 10²²

Total binding energy in 3g of copper = 2.86 × 10²² × 551.385 MeV = 1.57696 × 10^25 MeV ≈ 1.6 × 10^25MeV